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I read on Wikipedia that "An example scale showing power ratios x, amplitude ratios √x" [1], i.e. that the relationship between power and amplitude is squared.

How come? Have this something to do with the relationship of power and electric potential, i.e. P = U² / R?

Where

  • P = Power
  • U = Electric potential
  • R = Resistance

Some hints can be given on the same article but still do not explain why [3]

My particular interest is about what exactly it means with -3 dB on current probes [4].

Links

warpi
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  • Are you asking about the power transmission of a wave? Time averaged electrical power in components of AC circuits? DC power in resistor? Amplitude of what type of quantity? That will affect how we try to explain the relationship. In general, power is (usually) proportional to the square of some kinematic variation. The specific explanation can vary. – Bill N Nov 02 '20 at 20:10
  • @BillN In my particular case it was for current probes in electronics where it is written -3 dB. I never understood if that was power of amplitude... until now. I think the explanation from Claudio below is excellent. Goes from the original intent of decibel to a more broad description which makes intuitive sense. – warpi Nov 05 '20 at 18:19

2 Answers2

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Consider the function $P(A)$ that gives the power as a function of amplitude. We must have $P(0)=0$ and $P(A)>0$ for nonzero $A$. If $P$ is a well-behaved function, then we expect it to converge to its Taylor series near $A=0$. Generically, this means that $P(A)$ is of the form $(\text{const})A^2+\ldots$, where ... represents higher-order terms. It's conceivable that the constant in front of the $A^2$ could be zero, so that the leading-order dependence would be $A^4$, but this is unlikely and doesn't occur in any real-world examples that I know of.

It is not necessarily exactly true that $P\propto A^2$. This is not exactly true, for example, in the case of waves on a violin string. But the argument above shows why we expect it to be a good approximation when $A$ is small.

Urb
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user278751
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  • McLauren series of $P(A) =P(0)+P'(0)A+P"(0)A^2+P'''(0)A^3+\ldots$. Why did you drop the $P'(0)A$ term? – Bill N Nov 02 '20 at 19:49
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The wiki article about decibels shows a relation between amplitude and power.

As decibels was originally created to measure sound intensity, I understand it as amplitudes of sound waves x power of sound waves.

The sound waves can be modeled as a sum of trigonometric functions, each one of the form:

$\Delta x = A sin(kx - \omega t)$, where $\Delta x$ is the instantaneously longitudinal displacement of air molecules, and A is the amplitude of the oscillations.

The kinetic energy per unit of volume is $(1/2)\mu v^2$, where $\mu$ is the air density.

The velocity of the longitudinal displacements is the time derivative of $\Delta x$:

$\Delta v = -A\omega cos(kx - \omega t)$

The energy:

$E = (1/2)\mu v^2 = (1/2)\mu A^2\omega^2cos^2(kx - \omega t)$

As can be seen, energy is proportional to the square of the wave amplitude.

Claudio Saspinski
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